It will float because the density is less than 1
A model of what Shoulder pain looks like...
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The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.

Therefore,

In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,

Substitute the value of t from equation (1).

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

The shot put was thrown with a speed 15.02 m/s.
Answer:
the initial velocity is 20 m/s and the acceleration is 2 m/s²
Explanation:
Given equation of motion, v = 20 + 2t
If V represents the final velocity of the object, then the initial velocity and acceleration of the object is calculated as follows;
From first kinematic equation;
v = u + at
where;
v is the final velocity
u is the initial velocity
a is the acceleration
t is time of motion
If we compare (v = u + at) to (v = 20 + 2t)
then, u = 20 and
a = 2
Therefore, the initial velocity is 20 m/s and the acceleration is 2 m/s²